1. CRIMINAL RATIONAL EQUATIONS WHICH ARE REDUCED TO SQUARE EQUATIONS
An equation in which the variable is contained in the denominator of a fraction is called a fractional rational equation.
When solving fractional rational equations, most often, the procedure is:
– the denominators of multipliers are decomposed;
– the definition area of the equation is determined;
– we get rid of the fractions by multiplying the equation by NHS for the denominators;
– the obtained equation is solved;
– it is determined which of the obtained solutions are solutions of the initial equation.
Example 2. Let's solve the equation
The defining area of the equation is the set
D = R {-2, 2}.
Multiply the equation by (x – 2) (x + 2), so we get the equation:
х (х + 2) – З (х – 2) = 8,
that is, the equation:
х² – х – 2 = 0
whose solutions are х₁ = -1 and х₂ = 2. Since 2 ∉ D, it follows that the equation has only one root x = –1.
2. BINOMINAL EQUATIONS
The equation of type
xⁿ – а = 0
is called the nth degree binomial equation.
With the c
hange it comes down to a simpler equation:
уⁿ – 1 = 0
Depending on the coefficient а and the parity of the number n, for the binomial equation х ра – а = 0, the following applies:
1) If a = 0, the equation in the set of real numbers, ie the set of complex numbers has a unique solution x = 0
2) If a ≠ 0, then in the set of real numbers the equation for:
a) n = 2k and a> 0 has two real roots
b) n = 2k and a <0 has no real roots;
c) n = 2k + 1 and (∀a ∈ R) has a single real root
3) If a ≠ 0 is an arbitrary real number (or complex) number, then there are exactly n roots in the set of complex numbers.
3. BIQUARD EQUATIONS
The equation of type
ах⁴ + bx² + c = 0
where a, b, c are real numbers and a ≠ 0 is called a biquadratic equation.
We solve the biquadratic equation with the help of the change χ² = у which reduces it to the quadratic equation:
ay² + by + c = 0
If the solutions of the equation ay² + by + c = 0 are у₁ and у₂, then based on the change х² = у we get the set of equations:
x² = y₁, x² = y₂
For у₁, у₂ ∈ R, the solutions of the biquadratic equation are:
The biquadratic equation
аx⁴ + bx² + c = 0, at c = 0 gets the type:
ax⁴ + bx² = 0 ie x² (ax² + b) = 0
whose solution comes down to solving two incomplete quadratic equations:
x² = 0 and ах² + b = 0.
If b = 0, then the biquadratic equation аx⁴ + bx² + c = 0 is reduced to a fourth degree binomial equation ах⁴ + с = 0.
For this equation, at ace> 0 all four roots are complex, and for ace <0 it has two real and two complex roots.
If b = c = 0, then the biquadratic equation has a quadratic root x = 0.
4. TRINOMAL EQUATIONS
The equation of type
ах²ⁿ + bxⁿ + c = 0 where a, b, c ∈ R and а ≠ 0 is called a trinomial equation.
The equation is of the 2nd degree and its solution with the shift
хⁿ = у
comes down to solving the quadratic equation ay² + by + c = 0.
If u₁ and u₂ are the roots of the last equation, then the roots of the trinomial equation are obtained by solving the set of binomial equations:
хⁿ = У₁
хⁿ = У₂
5. SQUARE EQUATIONS
These equations usually have the type:
a[ f (x)²] + b [ f (x)]+ c = 0
where a, b, c ∈ R (a ≠ O), and f (x) is an expression that depends on x.
Such equations are called quadratic equations, because the change f (x) = u comes down to a quadratic equation:
ay² + by + c = 0
One root of the third degree symmetric equation is always the number -1, and the other two are obtained by solving the quadratic equation:
ах² + (b – a) x + a = 0
The third degree equation of type
ах³ – bx² – bx – a = 0, (а ≠ 0)
is solved in a similar way as the third degree symmetric equation. The number 1 is always the solution to this equation.
6. FOURTH DEGREE SYMMETRIC EQUATION
The equation of type:
ах⁴ + bx³ + сх² + bx + а = о, (а ≠ 0), is called the fourth degree symmetric equation.
By grouping the terms with equal coefficients, we get the equation:
а (х⁴ + 1) + b (x³ + х) + сх² = 0
Divide this equation by x² (x ≠ 0) and get the equation
(*)
We introduce a new variable with the shi
ft, from which squaring is obtained:
that is
By substituting the corresponding values in (*), we get the pawn
а (у² – 2) + by + c = 0
If у₁, у₂ ∈ R are the roots of the last equation, then the values of x are found from the equations:
that is, from the set of equations:
Hence the four roots of the symmetric equation are obtained, provided that у₁, у₂ ∈ R
The quadratic equation of the type:
ax⁴ + bx³ ± bx – a = 0 (a ≠ 0),
is solved by decomposing its left-hand side into factors, giving:
(х² ± 1) (ах² + bx + a) = 0
7. FOURTH DEGREE SYMMETRIC EQUATIONS
The equation of form:
ах⁵ + bx⁴ + сх³ + cx² + bx + а = 0, (a. 0) is a fifth degree symmetric equation.
The solution procedure is as follows:
– the members with the same coefficients are grouped:
a (x³ + 1) + bx (x³ + 1) + cx² (x + 1) = 0
* x³ + 1 = (x + 1) (x² – x + 1) and
* x⁵ + 1 = (x + 1) (x⁴ – x³ + x² – x + 1)
the equation takes the form:
(x + 1)[ ax⁴ + (b-a)x³ + (a – b + c)x² + (b-a) x + a] = 0
One root is x₁ = -1, and the other 4 roots are obtained by solving the fourth degree symmetric equation:
ax⁴ + (b-a) x³ + (a – b + c) x² + (b-a) x + a = 0
The fifth degree equation of the type:
ах⁵ + bx⁴ + сх³ + cx² – bx – а = 0, (a ≠ 0),
is solved in a similar way as the fifth degree symmetric equation. One solution to this equation is x = 1. This equation is called the fifth degree obliquely symmetric equation.
– equations that are squared –